Question

# If the value of ${{i}^{2}}=-1$, then calculate the value of $3{{i}^{2}}+{{i}^{3}}-{{i}^{4}}$.(a) $-4-i$(b) $-2-i$(c) $2+i$(d) $4+i$(e) $6+2i$

Hint: To solve this question we will assume variables a, b, c to $3{{i}^{2}},{{i}^{3}}$ & $-{{i}^{4}}$ and we will use the identity that ${{i}^{2}}=-1$. Finally we will add them up to get the result.

Complete step-by-step solution:
We are given that the value of ${{i}^{2}}=-1$, then the value of ${{i}^{3}}$, ${{i}^{4}}$ can be calculated separately.
Firstly we will calculate the value of ${{i}^{3}}$ and ${{i}^{4}}$, thus proceed to calculate the value of $3{{i}^{2}}+{{i}^{3}}-{{i}^{4}}$.
We have, ${{i}^{2}}=-1$.
Then, Let $a=3{{i}^{2}},b={{i}^{3}},c=-{{i}^{4}}$.
We have to calculate the value of a + b + c,
Because, ${{i}^{2}}=-1$.
\begin{align} & \Rightarrow 3{{i}^{2}}=\left( 3 \right)\left( -1 \right) \\ & \Rightarrow 3{{i}^{2}}=-3 \\ \end{align}
Therefore, $a = -3$ ------- (1)
Now consider b.
$b={{i}^{3}}$
We have, ${{i}^{2}}=-1$.
Multiplying āiā both sides of the above equation,
$\Rightarrow {{i}^{3}}=-i$
$\Rightarrow b = -i$ ---------- (2)
Now compute, $c=-{{i}^{4}}$.
We have, ${{i}^{2}}=-1$.
Multiplying, ${{i}^{2}}=-1$ on both sides we have,
\begin{align} & {{i}^{4}}=\left( -1 \right)\left( -1 \right) \\ & \Rightarrow {{i}^{4}}=1 \\ \end{align}
Now, $c=-{{i}^{4}}=-1$.
Hence, $c = -1$ ā------ (3)
Now a + b + c, using (1), (2) & (3) we have,
$a+b+c=-3-i-1=-4-i$, which is option (a).
Therefore, $3{{i}^{2}}+{{i}^{3}}-{{i}^{4}}=-4-i$, option (a) is correct.

Note: Another way to solve this question can be directly. Substituting, ${{i}^{2}}=-1$, ${{i}^{4}}=1$ & ${{i}^{3}}=-i$ to get the result, then answer would come as $3{{i}^{2}}+{{i}^{3}}-{{i}^{4}}=3\left( -1 \right)+\left( -i \right)-1=-4-i$, option (a).